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princessa240898
@princessa240898
July 2022
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sin3x-2cos2x-sinx=0 помогите пожалуйста))
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ribinaolga
Sin3x-2cos2x-sinx=0
(sin3x-sinx)-2cos2x=-
2sinxcos2x-2cos2x=0
2cos2x(sinx-1)=0
cos2x=0 sinx-1=0
2x=π/2+πn,n∈Z sinx=1
x=π/4+πn/2, n∈Z x=π/2+2πk, k∈Z
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Answers & Comments
(sin3x-sinx)-2cos2x=-
2sinxcos2x-2cos2x=0
2cos2x(sinx-1)=0
cos2x=0 sinx-1=0
2x=π/2+πn,n∈Z sinx=1
x=π/4+πn/2, n∈Z x=π/2+2πk, k∈Z